Let $a$ and $b$ respectively be the points of local maximum and local minimum of the function $f\left(x\right)=2x^3-3x^2-12x$. If $A$ is the total area of the region bounded by $y=f\left(x\right),$ the $x$-axis and the lines $x=a$ and $x\mathit{=}b\mathit{,}$ then $4\mathrm{A}$ is equal to ______.

.$f^{\text{'}}\left(x\right)=6x^2-6x-12=6\left(x^2-x-2\right)$

$=6(x-2)(x+1)$

Clearly, $f^{\text{'}}\left(-1^{-}\right)>0 \& f^{\text{'}}\left(-1^{+}\right)<0$

Hence, $x=-1$ is the point of local maximum.

Also, $f^{\text{'}}\left(2^{-}\right)<0 \& f^{\text{'}}\left(2^{+}\right)>0$

Hence, $x=2$ is the point of local minimum.

$\therefore a=-1 \& b=2$

Points=$\left(2,-20\right) \& \left(-1,7\right)$

$A={\int }_{-1}^0\left(2x^3-3x^2-12x\right)dx-{\int }_0^2\left(2x^3-3x^2-12x\right)dx$

$={\left(\frac{x^4}{2}-x^3-6x^2\right)}_{-1}^0-{\left(\frac{x^4}{2}-x^3-6x^2\right)}_0^2$

$$\begin{gathered} =\frac{9}{2}+24=\frac{57}{2} \\ \Rightarrow 4 \mathrm{A}=114 \end{gathered}$$